c++ - 具有混合继承修饰符的菱形继承(钻石问题)( protected /私有(private)/公共(public))

Question

假设我们有 类 A、B、C、D，其中 A 是基础，B、C 是介于两者之间，D 是在菱形模型中派生的。

注意:

class B 在private 模式下继承virtualy class A，

C 类 在保护 模式下继承虚拟 A 类。

class A
{
public:
    int member;  // note this member
};
class B :
    virtual private A // note private 
{

};
class C :
    virtual protected A // note protected
{

};
class D :
    public B, // doesn't metter public or whatever here
    public C
{

};

int main()
{
    D test;
    test.member = 0; // WHAT IS member? protected or private member?
    cin.ignore();
    return 0;
}

现在当我们创建一个 class D 的实例时，member 会是什么？私有(private)或 protected 哈哈？

图 2:

如果我们这样做会怎样:

class B :
    virtual public A // note public this time!
{

};
class C :
    virtual protected A // same as before
{

};

我想 member 将在第二个示例中公开，对吗？

Answer 1

§11.6 多路访问 [class.paths]

If a name can be reached by several paths through a multiple inheritance graph, the access is that of the path that gives most access. [ Example:

class W { public: void f(); };
class A : private virtual W { };
class B : public virtual W { };
class C : public A, public B {
   void f() { W::f(); } // OK
};

Since W::f() is available to C::f() along the public path through B, access is allowed. —end example ]

我想我不需要添加任何其他内容，但另请参阅 this defect report (被关闭为“不是缺陷”)。